If A = <2 ,1 ,6 >A=<2,1,6>, B = <5 ,1 ,3 >B=<5,1,3> and C=A-BC=A−B, what is the angle between A and C?

1 Answer
Jim G.
Jan 14, 2016

< theta = 63.8^@ θ=63.8∘ ( 1 decimal place )

Explanation:

vecC = vecA - vecB =( 2 , 1 ,6 ) - (5 , 1 ,3 ) =( - 3 , 0 , 3 ) →C=→A−→B=(2,1,6)−(5,1,3)=(−3,0,3)

To calculate the angle (theta) between vecA and vecC (θ)between→Aand→C

use the following formula :

costheta =( vecA . vecC)/ (|vecA| |vecC|)cosθ=→A.→C∣∣∣→A∣∣∣∣∣∣→C∣∣∣

The 'dot product' vecA . vecC =(2 , 1 , 6 ) . ( - 3 , 0 , 3 )→A.→C=(2,1,6).(−3,0,3)

= -6 + 0 + 18 = 12

and |vecA| = sqrt(2^2 +1^2 + 6^2) = sqrt(4 + 1 + 36 ) = sqrt41∣∣∣→A∣∣∣=√22+12+62=√4+1+36=√41

|vecC| = sqrt( (- 3 )^2 +0^2 + 3^2) = sqrt(9 + 0 + 9) = sqrt18 ∣∣∣→C∣∣∣=√(−3)2+02+32=√9+0+9=√18

substituting these values into the formula :

costheta = 12/(sqrt41 xx sqrt18) = 0.4417...

rArr theta = 63.8^@

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